# How does $\dfrac{f(x)}{g(x)}\approx \dfrac{f'(a)(x-a)+f(a)}{g'(a)(x-a)+g(a)} \implies \dfrac{f'(a)}{g'(a)}$?

Thus $$\frac{f(x)}{g(x)}\approx \color{red}{\frac{f'(a)(x-a)+f(a)}{g'(a)(x-a)+g(a)}}.$$ Taking the limit of the right hand side gives $$\dfrac{f'(a)}{g'(a)}$$.

Because $$x \to a \iff x- a \to 0$$, then

$$\lim_{x \to a} \color{red}{\dfrac{f'(a)(x-a)+f(a)}{g'(a)(x-a)+g(a)}} = \dfrac{f'(a)\times0 + f(a)}{g'(a)\times0 + g(a)}$$

Now what?

• In that reference, one is applying this to the case $f(a)=f(b)=0$ (in order to justify the Hospital). – Angina Seng Jan 12 '19 at 3:29

If $$g(a)\ne0$$, we can just write $$\frac{f(x)}{g(x)}\approx\frac{f(a)}{g(a)}$$.

If $$f(a)\ne0=g(a)$$, $$\frac{f(x)}{g(x)}$$ explodes to $$\pm\infty$$.

But if $$f(a)=g(a)=0\ne g^\prime(a)$$, that's more interesting.

In fact, your question contains almost all the logic we need for it: substituting $$f(a)=g(a)=0,\,x\ne a$$ gives$$\frac{f(x)}{g(x)}\approx\frac{f^\prime(a)(x-a)+0}{g^\prime(a)(x-a)+0}=\frac{f^\prime(a)(x-a)}{g^\prime(a)(x-a)}=\frac{f^\prime(a)}{g^\prime(a)}.$$

What you mention is an attempt to prove the following simple result :

Simple Theorem: If functions $$f, g$$ are differentiable at $$a$$ and $$f(a)=g(a) =0,g'(a)\neq 0$$ then $$\lim_{x\to a} \frac{f(x)} {g(x)} =\frac{f'(a)} {g'(a)}$$

The presentation in your question in not rigorous and the result is simply obtained using definition of derivative. We have $$\frac{f(x)} {g(x)} =\dfrac{\dfrac{f(x) - f(a)} {x-a}} {\dfrac{g(x) - g(a)} {x-a}}$$ and clearly as $$x\to a$$ the right hand side tends to $$f'(a) /g'(a)$$ via definition of derivative.

A smooth function is locally linear:

$$f(x)\approx f(a)+(x-a)f'(a)$$

and when $$f(a)=0$$, this simplifies to

$$f(x)\approx (x-a)f'(a).$$

L'Hospital easily follows:

$$f(a)=g(a)=0\implies \frac{f(x)}{g(x)}\approx \frac{f'(a)}{g'(a)}$$

and the approximation becomes more and more accurate as $$x\to a$$.

More rigorously,

$$f(x)=f(a)+(x-a)f'(a)+(x-a)r(x,a)$$ where $$r$$ tends to $$0$$ when $$x$$ tends to $$a$$.

Then

$$\frac{f(x)}{g(x)}=\frac{f'(a)+r(x,a)}{g'(a)+s(x,a)}$$ and

$$\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(a)+r(x,a)}{g'(a)+s(x,a)}=\frac{f'(a)}{g'(a)}.$$